integral from-2 to 1 of x(4-x^2)-(x+2)

解

\int_{- 2}^{1} x (4 - x^{2}) - (x + 2) d x

- \frac{27}{4}

十進法表記

- 6.75

解答ステップ

\int_{- 2}^{1} x (4 - x^{2}) - (x + 2) d x

拡張

x (4 - x^{2}) - (x + 2) : - x^{3} + 3 x - 2

= \int_{- 2}^{1} - x^{3} + 3 x - 2 d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 2}^{1} x^{3} d x + \int_{- 2}^{1} 3 x d x - \int_{- 2}^{1} 2 d x

\int_{- 2}^{1} x^{3} d x = - \frac{15}{4}

\int_{- 2}^{1} 3 x d x = - \frac{9}{2}

\int_{- 2}^{1} 2 d x = 6

= - (- \frac{15}{4}) - \frac{9}{2} - 6

簡素化

- (- \frac{15}{4}) - \frac{9}{2} - 6 : - \frac{27}{4}

= - \frac{27}{4}